Can any metric space be completed? Completion defined in Real Analysis, Carothers, 1ed has been captured below.

Can any metric space be completed?
 A: Yes. Rudin outlines a proof in Principles of Mathematical Analysis, Chapter 3, Problem 24.
Given a metric space $(X,d)$, say two Cauchy sequences $(p_n)$ and $(q_n)$ are equivalent if $\lim_{n\to\infty} d(p_n,q_n) = 0$. This is an equivalence relation on the set of all Cauchy sequences, as you can verify.
The set of all equivalence classes, say $X^*$, can be given a metric: if $P$ and $Q$ are equivalence classes in $X^*$ and $(p_n)\in P$, $(q_n)\in Q$ are representatives for the equivalence classes, set $\Delta(P,Q) = \lim_{n\to\infty} d(p_n,q_n)$. It can be shown that this is a metric on $X^*$. Moreover, $(X^*, \Delta)$ is a complete metric space, and there is an isometry of $X$ into $X^*$, so $(X^*, \Delta)$ is a completion of $(X,d)$ by Carothers's definition of the term.
This is an excellent exercise, and I highly encourage you to find a copy of Rudin and try it out. He gives more details and hints in his outline, so with substantial effort you should be able to give it a good go.
A: Cauchy sequences are a standard method of completing a metric space. An alternative method of completion of a metric space $(M, d)$ is given by the Kuratowski embedding into the space of bounded functions $M \to \mathbb R$. Exercise: prove that this space of bounded functions forms a complete metric space under the metric $\hat d(f, g) = \displaystyle\sup_{x \in M} |f(x)-g(x)|$. If we name this metric space $N$, and let $x_0$ be an arbitrary element of $M$, then $M$ embeds into $N$ by the map $x \mapsto \{y \mapsto d(x,y)-d(x,x_0)\}$. The purpose of subtracting $d(x,x_0)$ is simply to ensure the map is bounded. Exercise: prove this gives an isometric embedding. Then the completion $\hat M$ is just given by the closure of the image of this embedding.
